Chemistry 163 General Chemistry 3 Your Instructor is

Chemistry& 163 General Chemistry 3 Your Instructor is Dr Lee West Welcome !!

First of all lets get the formalities out of the way…. Instructor: Lee West Office: LSC-114 Phone: (253) 864 -3353 Email: lwest@pierce. ctc. edu Office Hours: M-Thurs 10 AM– 10: 50 AM and 1 PM-2 PM

Class Times: Mon-Thurs 11: 00 AM – 11: 50 AM (CTR 178) Lab Time: Fri 10 -11: 50 AM (7055) or 12 -1: 50 PM (7056) (LSC 103) Textbook: Chemistry, Openstax College https: //www. openstaxcollege. org/textbooks/chemistry/get Calculator: Students must have a basic scientific, calculator. It is not acceptable to use a mobile phone as your calculator. “Sapling Learning” access “Carbonless Copy” Bound Lab Notebook

Quizzes: Short quizzes will be given in class on Thursdays when no test is scheduled. No makeup quizzes will be given, but the lowest quiz score will be dropped.

Homework: Homework from the book for each chapter will be peer evaluated. Online homework will be assigned using Sapling Learning.

Tests: Three tests will be given. Make-up tests will only be allowed for excused absences (e. g. illness with a doctor’s note). Final Exam: The final exam is comprehensive and will cover all material presented during the general chemistry sequence.

Laboratory requirements: Experiments require up to 1 hr 50 mins. It is not possible to make up laboratory classes. You must maintain a 75 % attendance in order to participate in the laboratory exercises.

Chapter 12: Kinetics 12. 1 Chemical Reaction Rates 12. 2 Factors Affecting Reaction Rates 12. 3 Rate Laws 12. 4 Integrated Rate Laws 12. 5 Collision Theory 12. 6 Reaction Mechanisms 12. 7 Catalysis

Chapter 12: Kinetics 12. 1 Chemical Reaction Rates

A spontaneous process can be used to drive a nonspontaneous process. How fast the two processes are occurring is also very important.

Consider our use of oil and gas reserves. Formation of our oil and gas reserves occurred over millions of years. Consumption is occurring much quicker.

In order to truly understand a chemical system we need to know: • The identity of the products and reactants • The stoichiometry • Energy changes occurring • Entropy changes occurring (spontaneity) • Chemical kinetics (how fast)

If we consider the formation of NO: N 2(g) + O 2(g) → 2 NO(g) The rate of change in the concentration of NO is given by: In chemistry square brackets indicates concentration (typically in mol/L).

In chemistry rates of reactions are always defined as positive numbers: N 2(g) + O 2(g) → 2 NO(g) The rate of change in the concentration of N 2 is given by: Similarly,

Consideration of the stoichiometry tells us NO will be produced at twice the rate O 2 and N 2 are consumed: N 2(g) + O 2(g) → 2 NO(g)

For any reaction the relationship between the rates of consumption of reactants and formation of products is: a. A + b. B → c. C + d. D

Example: The oxidation of carbon monoxide to carbon dioxide occurs as below: 2 CO(g) + O 2(g) → 2 CO 2(g) (a) Which reactant is consumed at the greatest rate? (b) How is the rate of change in [CO 2] related to the change in [O 2]? (a) CO(g) is consumed at twice the rate of O 2 (b)

Example: Consider the reaction: A + 2 B → C When the [B] is decreasing at 0. 500 mol/(Ls) how fast is the [A] decreasing?

Example: The relationship between the reaction rates of a system is found to be: Write a balanced equation for the reaction. CH 4 + 2 O 2 → 2 H 2 O + CO 2

Example: Consider the following reaction: 2 NO(g) + O 2(g) → 2 NO 2(g) If the rate of consumption of O 2 is 0. 022 mol/(Ls) what is the rate of formation of NO 2?

Consider the following data for the formation of NO from N 2 and O 2. N 2(g) + O 2(g) → 2 NO(g) The reaction rate changes with time!

We can determine the average rate of formation of NO between 5. 0 ms and 10. 00 ms.

Likewise we can determine the average rate of formation of NO between 5. 0 ms and 15. 00 ms.

We may also determine the instantaneous rate of a reaction. The instantaneous rate of a reaction is determined by drawing a line tangent to the curve of concentration versus time and determining its slope.

The instantaneous rate of a reaction is quite awkward to determine (practice!!). You need to plot a few points either side of the time of interest then draw in the tangent line and determine its slope.

Chapter 12: Kinetics § 12. 1 Chemical Reaction Rates Using a balanced equation it is possible to determine the relative rates of consumption of reactants and formation of products. The rate of a chemical reaction can be expressed quantitatively as the average rate over a particular period of time or the instantaneous rate at a single time.

Chapter 12: Kinetics 12. 2 Factors Affecting Reaction Rates

What must happen for a reaction to occur? Reactants must collide • Anything that causes an increase in collisions between the reacting particles will increase the rate. Increasing concentration tends to lead to an increase in reaction rate

As a reaction proceeds the rate of consumption of the reactants decreases. The reaction rate is fastest when t = 0. This is referred to as the initial rate.

This makes intuitive sense, in order for a reaction to occur the reactants must come into contact. The more reactants in a given space the greater the likelihood of contact.

Solids interact through their surfaces, finely dividing a solid increases surface area and the rate of reaction. e. g. Granulated sugar dissolves faster than lumps. “dust explosions” result from suspension of finely divided powders in air. (Extreme industrial hazard)

Increasing the temperature increasing the “average” speed of the particles and the likelihood of them colliding. This results in an increased reaction rate. We know this from experience, If we want to dissolve a solid in a liquid we can warm the liquid.

When reactants collide they must do so with enough energy to break bonds in the reactants and allow bonds to form in products. e. g. N 2 + O 2 → 2 NO : N N: : . O. : . N O:

The minimum energy that the reactants must have in order to allow the reaction to occur is called the activation energy, EA. As the temperature increases the proportion of reactants with energy > EA increases and the reaction goes faster.

Catalysts are species that speed up a reaction but are not consumed in the reaction. Catalysts provide an alternative reaction pathway of lower Ea. Ea catalyzed

Chapter 12: Kinetics 12. 2 Factors Affecting Reaction Rates Factors which affect the rate of a chemical reaction include the temperature, the concentration of reactants, the state of subdivision and the presence of a catalyst.

Chapter 12: Kinetics 12. 3 Rate Laws

Consider the reaction: A → B Concentration (mol. L-1) 0. 12 0. 1 0. 08 0. 06 [A] 0. 04 [B] 0. 02 0 0 10 Time (s) 20 30 As the reaction progresses the rate of the forward reaction will decreases and the rate of the reverse reaction will increases.

To avoid this complication it is common practice to calculate the “initial rate” i. e. the instantaneous rate of reaction when the reagents are mixed. Concentration (mol. L-1) 0. 12 0. 1 0. 08 0. 06 [A] 0. 04 [B] 0. 02 0 0 10 Time (s) 20 30 We will always use initial rates to determine kinetic information.

A key aim of most kinetic studies is to deduce the rate law which describes how the concentration of reactants and products varies with time. For a general reaction: a. A + b. B → c. C + The rate law has the form: Rate = k[A]m[B]n d. D

For the reaction: a. A + b. B → c. C + d. D Rate = k[A]m[B]n • k is the rate constant and is fixed for a given temperature. • The exponents m and n are the reaction orders of A and B, respectively. • m+ n = overall reaction order

How can we determine the reaction orders? In developing our strategy we can keep these points in mind. 1. The reaction orders need to be found experimentally and cannot be determined from the balanced equation. 2. Using initial rates avoids the complications of the reverse reaction.

A convenient way of determining the reaction orders is the “method of initial rates”. Consider the example reaction in our text book: O 2(g) + 2 NO(g) → 2 NO 2(g) The rate law for this reaction has the form: Rate = k[O 2]m[NO]n To find the reaction orders we run a series of experiments. Two with constant [O 2] and varying [NO] and two with constant [NO] and varying [O 2].

For each of these reactions we obtain the initial rate: Experiment Initial [O 2] Initial [NO] Initial Rate mol/(Ls) 1 1. 10 x 10 -2 1. 30 x 10 -2 3. 21 x 10 -3 2 1. 10 x 10 -2 2. 60 x 10 -2 12. 8 x 10 -3 3 2. 20 x 10 -2 1. 30 x 10 -2 6. 40 x 10 -3

Similarly,

We can now obtain k by substituting data from one of our experiments into the rate law: Determining the units for k is not always trivial.

Fortunately there is a “simple rule” to give the units of the rate constant. Overall Reaction Order Units of k m conc(1 -m) time-1 0 conc time-1 1 time-1 2 conc-1 time-1 3 conc-2 time-1

Chapter 12: Kinetics 12. 3 Rate Laws The rate law of a reaction describes the dependence of the rate of reaction on the concentration of reactants. The method of initial rates provides a convenient method for determining the rate law.

Chapter 12: Kinetics 12. 4 Integrated Rate Laws

One of the difficulties with the method of initial rates is it requires multiple experiments to be performed. e. g. 2 NO + O 2 → 2 NO 2 Experiment Initial [O 2] Initial [NO] Initial Rate mol/(Ls) 1 1. 10 x 10 -2 1. 30 x 10 -2 3. 21 x 10 -3 2 1. 10 x 10 -2 2. 60 x 10 -2 12. 8 x 10 -3 3 2. 20 x 10 -2 1. 30 x 10 -2 6. 40 x 10 -3

For reactions involving a single reactant it is possible to perform a single experiment and determine the rate law. e. g.

Consider the decomposition of O 3 in sun light: e. g. sunlight O 3 → O 2 + O 2 The reaction is 1 st order in O 3 and we can write:

Using the rules of integration we can determine: A plot of ln[O 3] versus time gives a straight line with slope –k and intercept [O 3]0

In general terms, for a first order reaction with one reactant: A → B + C + …. The integrated rate law is: A plot of ln[A] versus t will give a straight line with: • intercept = ln[A]0 • slope = -k.

Example: The following data was collected for the decomposition of N 2 O 5: 2 N 2 O 5 → 2 N 2 O 4 + O 2 (a) Is the reaction 1 st order wrt to N 2 O 5 ? Provide evidence for your answer. (b) What is the rate constant for the reaction ? A plot of ln[N 2 O 5] versus time gives a straight line with negative slope. Hence, the reaction is 1 st order wrt to N 2 O 5.

Example: The following data was collected for the decomposition of N 2 O 5: 2 N 2 O 5 → 2 N 2 O 4 + O 2 (a) Is the reaction 1 st order wrt to N 2 O 5 ? Provide evidence for your answer. (b) What is the rate constant for the reaction ? NOTE: k has units (time)-1 for a 1 st order process

A useful measure of how fast a reaction is occurring is the half-life, t 1/2, the time for the concentration of a reactant to halve. Consider a 1 st order decomposition reaction: A → B + C

For a 1 st order process t 1/2 is given by: For a 1 st order process the half-life is independent of the concentration:

Example: The rate constant for the decomposition of N 2 O 5 is 7. 8 x 10 -3 s-1. What is the half-life of N 2 O 5? The units of the rate constant tell us this is a 1 st order process. Hence, the half-life is given by:

For a reaction with one reactant : A → B + C + …. If the reaction is second order with respect to A then we would write: Using the rules of integration we can determine:

For a reaction that is 2 nd order with respect to A if we plot versus t we will get a straight line with: • intercept = 1/[A] • slope = k Slope = k Intercept = 1/[A]0 Time (s)

Example: Cl. O decomposes according to the reaction: 2 Cl. O → Cl 2 + O 2 The kinetics of the reaction were studied and the following data obtained. (a) Is the reaction 1 st or 2 nd order wrt to Cl. O? (b) What is the rate constant for the process? The reaction is 2 nd order

Example: Cl. O decomposes according to the reaction: 2 Cl. O → Cl 2 + O 2 The kinetics of the reaction were studied and the following data obtained. (a) Is the reaction 1 st or 2 nd order wrt to Cl. O? (b) What is the rate constant for the process? Note: for a 2 nd order process k has units (conc)-1(time)-1

We can derive an expression for the half-life of a 2 nd order process: The half-life of a 2 nd order process is dependent on the initial concentration.

Example: Calculate t 1/2 for the 2 nd order decomposition of NO 2: 2 NO 2 → 2 NO + O 2 given the rate constant is 0. 544 M-1 s-1 and the initial concentration of NO 2 is 0. 0100 M.

The integrated rate law for a 2 nd order process: Applies only in the case of a single reactant. e. g. A → B + C It does not apply in the case of a 2 nd order reaction that is 1 st order in two reactants. e. g. A + B → C Rate = k[A][B]

One approach to studying 2 nd order reactions that are 1 st order in two reactants is to have a very large concentration of one reactant. A + B → C Rate = k[A][B] Under these conditions the small change in the concentration of the excess reactant does not affect the rate.

Under these conditions the reaction is described as being “pseudo first order”. A + B → C

Example: The following data were obtained for the reaction of NO in a large excess of O 3: NO + O 3 → NO 2 + O 2 (a) Is the reaction pseudo 1 st order wrt to NO? Provide evidence for your answer. (b) If yes, what is the pseudo first order rate constant for the process? A plot of ln[NO] versus t is linear, hence, the reaction is pseudo first order in NO

Example: The following data were obtained for the reaction of NO in a large excess of O 3: NO + O 3 → NO 2 + O 2 (a) Is the reaction pseudo 1 st order wrt to NO? Provide evidence for your answer. (b) If yes, what is the pseudo first order rate constant for the process?

For some rare cases the rate of the reaction is independent of reactant concentration. These reactions are described as being zero order. e. g. A → C If the reaction is zero order with respect to A then we would write: Using the rules of integration we can determine:

For a reaction that is zero order with respect to A if we plot [A] versus t we will get a straight line with: • intercept =[A]0 • slope = -k Intercept = [A]0 [A] Slope = -k Time (s)

Using this integrated rate law we can determine an expression for t 1/2 of a zero order reaction. The integrated rate law is: After one half-life:

Summarizing this in one table: Rate Law Integrated Rate Law Plot for Straight Line Slope, Intercept Half-Life Zero Order First Order Second Order Rate = k [A] =-kt+[A]0 Rate = k[A] ln[A] =-kt+ln[A]0 Rate = k[A]2 1/[A] =kt+1/[A]0 [A] versus t ln[A] versus t 1/[A] versus t -k, [A]0 -k, ln[A]0 k, 1/[A]0/2 k (ln 2)/k 1/k[A]0

Chapter 12: Kinetics 12. 4 Integrated Rate Laws Integrated rate laws have been developed for zero, first and second order reactions. These laws allow the determination of a concentration at a particular time. These laws allow us to distinguish between 1 st, 2 nd and zero order reactions on the basis of the shape of the plot of concentration versus time. Using these laws it is possible to develop expressions for the half-life of a reactant.

Chapter 12: Kinetics 12. 5 Collision Theory

What must happen for a reaction to occur? Reactants must collide • Anything that causes an increase in collisions between the reacting particles will increase the rate. Increasing concentration tends to lead to an increase in reaction rate

Increasing the temperature increasing the “average” speed of the particles and the likelihood of them colliding. This results in an increased reaction rate. We know this from experience, If we want to dissolve a solid in a liquid we can warm the liquid.

When reactants collide they must do so with enough energy to break bonds in the reactants and allow bonds to form in products. e. g. N 2 + O 2 → 2 NO : N N: : . O. : . N O:

The minimum energy that the reactants must have in order to allow the reaction to occur is called the activation energy, EA. As the temperature increases the proportion of reactants with energy > EA increases and the reaction goes faster.

Svante Arrhenius (1859 -1927) determined that this increase in rate was due to an exponential dependence of the rate constant, k, on T. Where: k = rate constant A = frequency factor Ea= activation energy k R = ideal gas constant T= temperature in Kelvin T (K)

The frequency factor, A, takes into account that not all collisions (even if they have Ea) will be effective. e. g. O 3 + NO → O 2 + NO 2 The reactants must have the correct orientation for the reaction to progress.

When the reactants collide with the correct orientation and kinetic energy > Ea an “activated complex” can form. O 3 + NO “activated complex” O 2 + NO 2 In the activated complex the “old” bonds have begun to break and the “new” bonds are forming.

The “activated complex” is an energy transition state between reactants and products. “activated complex” NO + O 3 NO 2 + O 2

Comparing the energy profile of the forward reaction to the reverse reaction: NO + O 3 → NO 2 + O 2 → NO + O 3 The reverse reaction has a much higher Ea and this reaction proceeds very slowly i. e. the reaction favors the forward direction.

The Arrhenius equation can be used to determine the Ea of a reaction. Taking the natural logarithm of both sides: This equation has the form y = mx +b Where: y = lnk m = -Ea/R x = 1/T b = ln. A

The linear form of the Arrhenius equation indicates a plot of lnk versus 1/T will be a straight line with slope –Ea/R:

Example: The following data was collected for the decomposition of Cl. O(g): 2 Cl. O → Cl 2 + O 2 What is the activation energy?

A plot of lnk versus 1/T will have slope –Ea/R

The Arrehnius equation can also be written in a convenient “two -point form”: Consider a reaction that is performed at T 1 and T 2

Example: The following data was collected for the decomposition of N 2 O 5(g): k (s-1) T (K) 2. 14 x 105 658 3. 23 x 105 673 105 688 7. 03 x 105 703 4. 81 x (a) What is the activation energy? (b) What is the rate constant at 300 K? 13. 6 lnk 13. 1 12. 6 12. 1 11. 6 0. 001402 0. 001452 0. 001502 0. 001552 1/T (K-1)

Example: The following data was collected for the decomposition of N 2 O 5(g): k (s-1) T (K) 2. 14 x 105 658 3. 23 x 105 673 105 688 7. 03 x 105 703 4. 81 x (a) What is the activation energy? (b) What is the rate constant at 300 K? (Makes sense that k is smaller as T is much lower!)

Chapter 12: Kinetics 12. 5 Collision Theory In order for reactions to occur reactants must collide with sufficient energy to overcome the activation energy and collide with the correct orientation. The quantitative relationship between temperature and the rate constant is described by the Arrhenius equation:

Chapter 12: Kinetics 12. 6 Reaction Mechanisms

A balanced chemical equation describes the changes in amounts of products and reactants as a reaction progresses. e. g. The reaction: 2 A + 3 B → A 2 B 3 Can be read as 2 moles of A combines with 3 moles of B to produce 1 mole of A 2 B 3. A chemical equation does not describe the way in which the molecules, ions and atoms physically come together.

The reaction mechanism describes the interactions that occur between the various species as a reaction progresses. The mechanism is described by a series of elementary steps which when summed give the overall reaction: e. g. A + B → AB A + AB → A 2 B B + A 2 B → A 2 B 2 B + A 2 B 2 → A 2 B 3 2 A + 3 B → A 2 B 3

The products of the elementary reactions that do not appear in the overall equation are called intermediates. A + B → AB A + AB → A 2 B B + A 2 B → A 2 B 2 B + A 2 B 2 → A 2 B 3 2 A + 3 B → A 2 B 3 Intermediates are relatively short lived species and it may not be possible to isolate them.

The elementary steps describe actual molecular events. The molecularity of an elementary reaction is defined by the number of reactant particles. A + B → AB 2 particles (bimolecular) A + AB → A 2 B 2 particles (bimolecular) B + A 2 B → A 2 B 2 2 particles (bimolecular) B + A 2 B 2 → A 2 B 3 2 particles (bimolecular) 2 A + 3 B → A 2 B 3

The molecularity of an elementary step is determined by the number of particles involved in that step: • One particle ≡ unimolecular • Two particles ≡ bimolecular • Three particles ≡ termolecular Termolecular elementary steps are rarely observed as three particles must come together with correct orientation and energy. Elementary steps involving four particles have not been observed.

Lets consider the mechanism of the decomposition of NO 2(g): Overall reaction is: 2 NO 2 → 2 NO + O 2 The reaction mechanism has two elementary steps: (1) 2 NO 2 → NO + NO 3 » NO 3 → NO + O 2 Overall: 2 NO 2 → 2 NO + O 2 What is the molecularity of each step? What are the intermediate/s? bimolecular unimolecular

Each step in a reaction mechanism has its own transition state and associated Ea. Ea, step 1 > Ea, step 2 Ratestep 1 < Ratestep 2 2 NO 2 → NO + NO 3 (slow) NO 3 → NO + O 2 (fast) Hint: # of “peaks” = # of elementary steps

Example: Draw the energy profile for an exothermic reaction with a two step mechanism that has a fast initial step and slow second step. Potential Energy Transition State 2 Transition State 1 Ea, step 2 Ea, step 1 Intermediate DH Reactants Products Reaction Progress

Unlike for the overall equation, the reaction order and rate equation of an elementary step can be deduced from the equation coefficients. e. g. Elementary step Molecularity Rate Law A 2 → 2 A Unimolecular k[A 2] 2 A → A 2 Bimolecular k[A]2 A + B → AB Bimolecular k[A][B] 2 A + B → A 2 B Termolecular k[A]2[B] Note: For an elementary step the reaction order = molecularity

Usually, one of the elementary steps proceeds at a much slower rate than the others. (1) 2 NO 2 → NO + NO 3 (slow) (2) NO 3 → NO + O 2 (fast) Rate 1 = k 1[NO 2]2 Rate 2 = k 2[NO 3] Overall: 2 NO 2 → 2 NO + O 2 The slowest step is called the “rate determining step” as it determines the rate of the overall reaction. 2 NO 2 → 2 NO + O 2 Rate = k[NO 2]2

Consider the following hypothetical reaction mechanism: (1) A + B → C (fast) (2) C D → (slow) Rate 1 = k 1[A][B] Rate 2 = k 2[C] Overall: A + B → D The overall rate is given by: Rate = k 2[C] The difficulty is C is an intermediate and we have no way of knowing its concentration.

Can we find a way of expressing the [C] in terms of the [A] and [B]? (1) A + B → C (fast and reversible) Rate 1 = k 1[A][B] (2) C D Rate 2 = k 2[C] → (slow) Overall: A + B → D Products from the first step will build up while they wait to be consumed by the second. This results in an increase in the rate of the reverse of step 1.

Eventually the rate of the reverse of step 1 increases until it equals the rate of the forward reaction: A + B → C Rate 1 = k 1[A][B] C A Rate -1 = k-1[C] → This allows us to write: + B

e. g. for our hypothetical reaction mechanism: (1) A + B → C (fast) (2) C D → (slow) Overall: A + B → D The overall rate is given by: Rate 1 = k 1[A][B] Rate 2 = k 2[C]

Example: It is proposed that the mechanism for the decomposition of N 2 O 5 to NO 2 is: Step 1: 2 N 2 O 5 → N 4 O 10 fast and reversible Step 2: N 4 O 10 → N 2 O 3 + 2 NO 2 + O 3 slow (rds) Step 3: N 2 O 3 + O 3 → 2 NO 2 + O 2 fast Overall: 2 N 2 O 5 → fast 4 NO 2 + O 2 What is the rate law for the overall reaction based on the proposed mechanism? If the mechanism is correct the reaction will be 2 nd order in N 2 O 5

Consider the following reaction: NO 2 + CO → NO + CO 2 Experimentally it is determined that the rate law is: Rate = k[NO 2]2 The rate law tells us the following about the mechanism: 1. The RDS does not involve CO 2. The RDS is bimolecular in NO 2 3. There is at least one fast step involving CO

Example: Propose a mechanism for the following reaction that is consistent with the observed rate law: NO 2 + CO → NO + CO 2 Step 1: 2 NO 2 → NO 3 + NO Step 2: NO 3 + CO → NO 2 + CO 2 Overall: NO 2 + CO → NO + CO 2 Rate = k[NO 2]2 slow (RDS) Rate = k[NO 2]2 fast

Chapter 12: Kinetics 12. 6 Reaction Mechanisms Reactions can be described in terms of the elementary steps making up the reaction mechanism. The relative speed of each step is determined by the Ea of that step. The rate law of the slowest step provides the rate law for the overall reaction. In reactions where the RDS involves an intermediate we must find an expression for the intermediate in terms of the reactants.

Chapter 12: Kinetics 12. 7 Catalysis

Catalysts are species that speed up a reaction but are not consumed in the reaction. Potential Energy Catalysts provide an alternative reaction pathway of lower Ea. Ea (uncatalyzed) Ea (Catalyzed) DH Reactants Products Reaction Progress

A well known example is the catalysis of O 3 decomposition in the stratosphere by CFC’s

The natural decomposition of O 3 occurs via a two step mechanism: Step 1: O 3 → O 2 + O fast Step 2: O 3 + O → 2 O 2 slow Overall: 2 O 3 → 3 O 2 Rate = k[O 3] The RDS has an Ea = 17. 7 k. J/mol

Trichlorofluoro methane (freon-11, CCl 3 F, CFC-11) was once commonly used as a refrigerant. US production ceased in 1996 In the stratosphere CFC’s are exposed to UV radiation with energy to break C-Cl bonds: CCl 3 F h → Cl + CCl 2 F

Free Cl atoms are highly reactive and catalyze the breakdown of O 3: Step 1: Cl + O 3 → Cl. O + O 2 Step 2: Cl. O + O 3 → Cl + 2 O 2 Overall: 2 O 3 → 3 O 2 The Cl atoms consumed in step 1 are reformed in step 2 so are not used up in the overall reaction.

The Ea for the decomposition of O 3 in the presence of Cl is only 2. 2 k. J/mol versus 17. 7 k. J/mol for the normal process. The Cl catalyzed process occurs much quicker on account of the lower Ea As Cl is not consumed in the process each Cl atom can be used many times until it reacts with something other than O 3.

The Cl atoms produced by the photolysis of CFC’s are an example of a homogeneous catalyst. • Both the Cl atoms and the O 3 are in the same phase • Enzymes are a biological example of homogeneous catalysts.

Heterogeneous catalysts are in a different phase from the reactants, a well known example is the “catalytic converter” in cars. Catalytic converters are designed to maximize the contact area of the catalyst 2 NO → N 2 + O 2

Many catalysts function by holding the reactant/s in the most favorable orientation for reaction. 2 NO ↓ N 2 + O 2

Chapter 12: Kinetics 12. 7 Catalysis Catalysts are substances that are not consumed in a chemical reaction but increases the rate of a reaction by providing an alternative pathway of lower Ea. Catalysts can be classified as either homogeneous or heterogeneous depending on whether they are in the same of a different phase as the reactants, respectively. Many catalysts function by holding the reactant/s in a position that increases the probability of a successful interaction.